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  • Subject: Solve an argument.

    The argument:
    While building a mount for a band saw motor, a friend of mine started to argue over the trig involved to get the most tension in the belt, given a fixed weight of the motor and lack of interest in adding bricks.

    After over a day of arguing, I'm sick of arguing. So I though id let a forum of machinists/engineers/highly intelligent people do it for me.
    We have drawn a set of figures, and writing out conclusions on what actually supplies the belt tension.
    Each figure, has pivot points connecting each segment. The top segment is considered a belt/rope, and as such is always held in tension by the lower segment, a rigid bar.
    The two segments are assumed to be weightless, with only the weight affixed to the end to simplify things.
    The lower bar is allways the same length for the sake of simplicity and eliminating another variable.

    We where both given this picture to rate from highest belt tension to lowest:
    http://www.voxelsoft.com/spacesim/pulleys2.png

    My conclusion:
    Its a simple matter of leverage advantage.
    The smaller the angle A, The less the top belt would strech per inch of the motor droped
    Any time you trade off distance, you gain force.

    The lever nature can be more easily seen after altering it slightly, as shown in my picture, with the new line being fixed in angle to the bottom arm, but the pivot point of the fulcrum being the lower left point. Basicly a 'bent' lever
    http://www.voxelsoft.com/spacesim/pulleys3.png

    As such: Tension on the belt highest with a small angle A (Higher leverage advantage) and angle B to be horizontal (mass furthest out)
    Assuming angle B translated into a value of 1 when horizontal, and 0 when vertical
    And angle A translated into a value of infinity at 0 (parallel), and 1 when at 90 degrees. I don't actually feel like doing the proper trig for this but im just looking for a

    very rough answer here. I know there's various trig functions to compute the actual vector of forces here.

    The tension on the top arm (or belt) is similar to Weight * A * B

    As such, both angles are equally important in producing the highest belt tension. With 90 degrees of A, you only get a max weight of 1, With near 90 degrees of B, you get

    nearly no weight to multiply by leverage advantage.

    The figures with the most tension would be from highest to lowest:
    5
    6 tied with 7
    1, 3, 2, 4 all very close together, But very far off from 5, 6 and 7.

    As far as the 'primary source' of the tension, Yes B must not be near 90 degrees, but generally, A will preferably be small in a band saw motor mount, and multiply the force by
    many times.

    I showed him a physical model of the setup, Showed him the tension on a rope increases as angle A was reduced until he was unable to even lift a 10lb weight by pulling on the string. He still does not believe me.

    http://www.voxelsoft.com/spacesim/pulleys4.png
    Thats his explantion.
    Last edited by Black_Moons; 08-20-2010, 03:42 AM.
    Play Brutal Nature, Black Moons free to play highly realistic voxel sandbox game.

  • #2
    5 will have the most tension. Show him by putting a fishermans scale into the tensioned element.

    Comment


    • #3
      You would see we both agree on 5 if you read his.
      Its the rest we don't agree on.
      Play Brutal Nature, Black Moons free to play highly realistic voxel sandbox game.

      Comment


      • #4
        If you draw the vectors of the forces, it appears that the ratio between the force in the string and the gravitational force of the weight is:
        1/(cos(a-b)*sin(b)/cos(b)+sin(a-b))

        The ratio approaches infinity if a and b go to zero.

        Now, that should settle it

        Edit:
        And let's put a picture in for good measure. Grid is +/-90 degrees, 0 degrees is omitted to prevent division by zero.
        You can see the angle a is the dominant factor here.


        Igor
        Last edited by ikdor; 08-20-2010, 06:26 AM.

        Comment


        • #5
          It's number 5. But, you don't have a lot of adjustment for the belt.

          General solution:
          The added vectors of motor's gravitational force and the belt tension should go through the pivot of the motor's mount plate. But that would give you the least adjustment for the belt tension. To get more adjustment, the vector should be above the pivot (for obvious reasons).


          Nick

          Comment


          • #6
            Originally posted by MuellerNick
            It's number 5. But, you don't have a lot of adjustment for the belt.
            Agreed. One solution would be to increase the angle inside of #5 and add a tightening mechanism such as a bolt or spring.

            Comment


            • #7
              Black Moons -- the minds eye tells you it's the way your talking about, however, your friend may be correct, Your both right on #5 and then he may be on the rest of them in proper or closer to proper order (I don't have the time to check them all but 6 and 7 may not be next in line after 5 and 2 could be more like it...)

              You may be making the minds eye mistake of thinking the closer the top and bottom fixed links are the higher the ratio and although true its not the only case for creating maximum tension due to the effect of other elements in play that are responsible for creating a greater or lesser predetermined load to begin with, -- the lower link and its relationship to horizontal and the juncture where the two links meet and his description is very accurate, re- read it and really try to think what he's saying...
              Last edited by A.K. Boomer; 08-20-2010, 09:10 AM.

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              • #8
                I know you simplified this to make it easier but don't forget that this analysis applies to static condition. When running under load it will matter which direction the drive turns.

                Tight side over the top will tend to lift the motor more than tight side underneath. I say over top and underneath without knowing exactly what your drive looks like but the principle remains.

                Really need a more complex model if you wish to figure the numbers.

                Comment


                • #9
                  If you really are doing this your wasting your time. A free swinging motor will always have belt slip and when it starts or hits a load the motor will rise and loosen the belt.

                  Just put a rod between motor and saw frame as a solid adjustment devise and be done with it.

                  If all this is for entertainment then continue on.
                  It's only ink and paper

                  Comment


                  • #10
                    What Carl said. The belt tension must be fixed so the motor doesn't bounce.

                    Now, if this is a theoretical question disguised as a real problem then we may ignore all practical considerations such as having sufficient adjustment.

                    The rule of small angles predicates that you may approximate the change in sin or cosine as a strictly linear function when either is close to zero. The error is negligible when the angles are less than a few degrees from the quadrants.

                    However, in this particular case it is the tiny changes in the sine and cosine that make all the difference. What we have is an inverse case of the rule of small angles. Since the position of the inner point (left) on the upper chord is approaching the centre in figure 5 it is beginning to exhibit the nearly linear rate of change of the sine of the angle=zero that the rule of small angles is based upon. On the other hand, the other end is approaching cosine=1 at an eponential rate.

                    The ratio then is changing in favour of angle b over angle a eponentially making angle b by far the predominant factor.
                    Free software for calculating bolt circles and similar: Click Here

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                    • #11
                      Belt Tension

                      You don't want the belt as tight as you can get it anyway. That puts extreme wear on the bearings of the motor and the spindle. As mentioned above, you want a rigid adjustment so you can make the belt tight enough not to slip, and maintain that tension, which is usually a long way from being stretched hard.

                      Comment


                      • #12
                        Originally posted by Carld
                        If you really are doing this your wasting your time. A free swinging motor will always have belt slip and when it starts or hits a load the motor will rise and loosen the belt.

                        Just put a rod between motor and saw frame as a solid adjustment devise and be done with it.

                        If all this is for entertainment then continue on.
                        I have to disagree with that. The accompanying photo shows a similar situation with a motor/gearbox combo hinged on one side and I have never experienced any belt slippage.

                        Strokersix has it right, the direction of rotation can either add or subtract belt tension. I always wondered if the size of the sheave effected the value of this component??

                        Dave

                        Last edited by becksmachine; 08-20-2010, 01:29 PM. Reason: forgot photo

                        Comment


                        • #13
                          We don't care about the pratical implimentations of this, Just the arguement of what has the most belt tension.

                          my friend does not believe ikdor's graph and wants further explination of the formual.

                          We actualy did plan to use a tension bolt: Did not solve the arguement of what has the most tension.
                          Last edited by Black_Moons; 08-20-2010, 01:58 PM.
                          Play Brutal Nature, Black Moons free to play highly realistic voxel sandbox game.

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                          • #14
                            The force applied equals ( the cosine of b times the cotangent of b) over the sine of a. It reaches a singularity at (angle = zero) = (force= infinity).
                            Free software for calculating bolt circles and similar: Click Here

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                            • #15
                              Hmm, I don't understand the cotangent in that equasion, Can't B only supply 1 mass? at 10 degrees for example, Cos(10)*(1/tan(10)) = 5.5851
                              (Found definition of cotangent as 1/tan)
                              Using windows calculator here in degree mode..

                              Seems like just the cos(10) = 0.9848 is the right amount of mass..
                              Play Brutal Nature, Black Moons free to play highly realistic voxel sandbox game.

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